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Multiply two eight bit numbers with shift and add method

Published on Nov 17 2009 // Trainees

Statement: Multiply the 8-bit unsigned number in memory location 2200H by the 8-bit unsigned number in memory location 2201H. Store the 8 least significant bits of the result in memory location 2300H and the 8 most significant bits in memory location 2301H.




Sample problem:

(2200) = 1100 (0CH)

(2201) = 0101 (05H)

Multiplicand = 1100 (1210)

Multiplier =  0101 (510)

Result = 12 x 5 = (6010)

Source program :

  • LXI H, 2200 : Initialize the memory pointer
  • MOV E, M : Get multiplicand
  • MVI D, 00H : Extend to 16-bits
  • INX H : Increment memory pointer
  • MOV A, M : Get multiplier
  • LXI H, 0000 : Product = 0
  • MVI B, 08H : Initialize counter with count 8
  • MULT: DAD H : Product = product x 2
  • RAL
  • JNC SKIP : Is carry from multiplier 1 ?
  • DAD D : Yes, Product =Product + Multiplicand
  • SKIP:  DCR B : Is counter = zero
  • JNZ MULT : no, repeat
  • SHLD 2300H : Store the result
  • HLT : End of program

Flowchart for program

35-Multiply two eight bit numbers with shift and add method 

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